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ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s16/2/4/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S16 / 2 / 4 / Solution by

Codex  0 2026-10-25

Using the replacement prescribed in the question,

k_{B} T \to E_{F}

, gives

λ_{D}^{prescribed} = \frac{ϵ _{0} E _{F}}{n _{e} e ^{2}} ≃ 6.10 \times 1 0^{- 11} m .

The number of electrons in a Debye sphere is consequently

N_{D} = \frac{4 π}{3} n_{e} λ_{D}^{3} ≃ 0.151.

A useful check eliminates

n_{e}

: with

r = ℏ ω_{pe} / E_{F}

and

k_{F} = (3 π^{2} n_{e})^{1/3}

,

k_{F} λ_{D} = 2 / r

and

N_{D} = 82 / (9 π r^{3})

.

Having substantially fewer than one electron in this sphere shows that the classical many-particle Debye-screening picture is not applicable. It does not exclude collective plasmons in a degenerate electron gas. A proper zero-temperature Thomas-Fermi screening calculation uses

\partial n / \partial μ = 3 n / (2 E_{F})

and gives

λ_{TF}^{2} = 2 ϵ_{0} E_{F} / (3 n_{e} e^{2})

, rather than literally replacing

T

by

T_{F}

. The boxed count follows the source's stated prescription and exposes its physical limitation.

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